Question

[Nguyễn Bảo Quyên]

Tính tổng: \(D=101^2+102^2+103^2+\cdots+200^2\)

Answer 1

[Pencil]

\(D=101^2+102^2+103^2+\cdots+200^2\)

\(D=101.\left(102-1\right)+102.\left(103-1\right)+103.\left(104-1\right)+\cdots+200.\left(201-1\right)\)

\(D=\left(101.102+102.103+103.104+\cdots+200.201\right)-\left(101+102+103+\cdots+200\right)\)

Đặt \(\left(101.102+102.103+103.104+\cdots+200.201\right)=A\)

\(3A=\left(103-100\right).101.102+\left(104-101\right).102.103+\left(105-102\right).103.104+\cdots+\left(202-199\right).200.201\)

\(3A=101.102.103-100.101.102+102.103.104-101.102.103+103.104.105-102.103.104+\cdots+200.201.202-199.200.201\)

\(3A=-\left(100.101.102\right)+200.201.202\) \(\)

\(3A=-1030200+8120400\)

\(3A=7090200\)

\(A=2363400\)

Đặt \(\left(101+102+103+\cdots+200\right)=B\)

Số số hạng của dãy B là: \(\frac{\left(200-101\right)}{1}+1=100\) (số hạng)

\(B=\frac{\left(101+200\right).100}{2}=15050\)

\(D=A-B=2363400-15050=2348350\)

Vậy tổng dãy \(D=2348350\)

Answer 2

[Nguyễn Thị Thương Hoài]

D = 101\(^2\) + 102\(^2\) + ... + 200\(^2\)

D = 101.(100 + 1) + 102.(101 + 1) + ...+ 200.(199 + 1)

D = 101.100 + 101 + 102.101 + 102 + ...+ 200.199 + 200

D = (100.101 + 101.102 +... + 199.200) + (101 + 102+...+200)

Đặt B = 100.101 + 101.102 + ...+ 199.200

C = 101 + 102 + ... + 200

100.101.3 = 100.101.(102 - 99) = 100.101.102-99.100.101

101.102.3 = 101.102.(103 - 100)= 101.102.103 - 100.101.102

............................................................................................................................

199.200. 3 = 199.200.(201-198) = 199.200.201 - 198.199.200

Cộng vế với vế ta có:

3.B = 199.200.201 - 99.100.101

3.B =7999800 - 999900

3B = 6999900

B = 6999900 : 3

B = 2333300

C = 101 + 102 + 103 + ... + 200

C = (200 + 101) x { [(200 -101) : 1 + 1] : 2}

C = 301 x {100 : 2}

C = 301 x 50

C = 15050

D = B + C

D = 2333300 + 15050

D = 2348350